CBSE Class 12 Mathematics Relations And Functions Class Test Set 11

Question 1. Let \(f: \{2, 3, 4, 5\} \to \{3, 4, 5, 9\}\) and \(g: \{3, 4, 5, 9\} \to \{7, 11, 15\}\) be functions defined by \(f(2) = 3\), \(f(3) = 4\), \(f(4) = 5\), \(f(5) = 5\), \(g(3) = g(4) = 7\) and \(g(5) = g(9) = 11\). Find \(gof\).
Answer: To find the composite function \(gof\), we need to evaluate \(g(f(x))\) for each element \(x\) in the domain of \(f\), which is \(\{2, 3, 4, 5\}\).
For \(x = 2\):
\((gof)(2) = g(f(2)) = g(3) = 7\)
For \(x = 3\):
\((gof)(3) = g(f(3)) = g(4) = 7\)
For \(x = 4\):
\((gof)(4) = g(f(4)) = g(5) = 11\)
For \(x = 5\):
\((gof)(5) = g(f(5)) = g(5) = 11\)
Thus, \(gof = \{(2, 7), (3, 7), (4, 11), (5, 11)\}\).
In simple words: The composite function \(gof\) is found by applying function \(f\) first and then feeding its output into function \(g\).

Exam Tip: Always verify that the range of the inner function is a subset of the domain of the outer function before finding the composite.

Question 2. If \(f: R \to R\) is given by \(f(x) = (3 - x^3)^{\frac{1}{3}}\). Find \((fof)(x)\).
Answer: Given \(f(x) = (3 - x^3)^{\frac{1}{3}}\).
We need to find \((fof)(x) = f(f(x))\):
\((fof)(x) = f\left((3 - x^3)^{\frac{1}{3}}\right)\)
Substitute \(f(x)\) into the function definition:
\((fof)(x) = \left(3 - \left((3 - x^3)^{\frac{1}{3}}\right)^3\right)^{\frac{1}{3}}\)
Simplify the cube term:
\((fof)(x) = \left(3 - (3 - x^3)\right)^{\frac{1}{3}}\)
\((fof)(x) = (3 - 3 + x^3)^{\frac{1}{3}}\)
\((fof)(x) = (x^3)^{\frac{1}{3}} = x\)
In simple words: Applying the function twice in this case brings you right back to the original input \(x\).

Exam Tip: When dealing with fractional powers, always check if powers multiply correctly, such as \((a^m)^n = a^{mn}\).

Question 3. Let \(f: N \to Y\) be a function defined by \(f(x) = 4x + 3\) where \(Y = \{y \in Y: y = 4x + 3 \forall x \in N\}\). Show that \(f\) is invertible. Find the inverse.
Answer: Step 1: Show that \(f\) is one-one (injective).
Let \(x_1, x_2 \in N\). Consider:
\(f(x_1) = f(x_2)\)
\(4x_1 + 3 = 4x_2 + 3\)
\(4x_1 = 4x_2 \implies x_1 = x_2\)
Thus, \(f\) is one-one.
Step 2: Show that \(f\) is onto (surjective).
By definition of the codomain, \(Y = \{y \in Y: y = 4x + 3 \forall x \in N\}\), which means every element in \(Y\) has a pre-image in \(N\). Thus, \(f\) is onto.
Since \(f\) is both one-one and onto, \(f\) is invertible.
Step 3: Find the inverse of \(f\).
Let \(y = f(x)\), so \(y = 4x + 3\).
Solve for \(x\) in terms of \(y\):
\(y - 3 = 4x\)
\(x = \frac{y - 3}{4}\)
Thus, the inverse function is \(f^{-1}(y) = \frac{y - 3}{4}\) or \(f^{-1}(x) = \frac{x - 3}{4}\).
In simple words: To prove invertibility, show that every output comes from one and only one input, and then swap \(x\) and \(y\) to find the inverse formula.

Exam Tip: To find the inverse quickly, set \(y = f(x)\) and make \(x\) the subject of the formula.

Question 4. If \(f(x) = \frac{4x + 3}{6x - 4}\), \(x \neq \frac{2}{3}\). Show that \((fof)(x) = x\) \(\forall x \neq \frac{2}{3}\). What is the inverse of \(f\)?
Answer: Step 1: Show that \((fof)(x) = x\).
\((fof)(x) = f(f(x)) = f\left(\frac{4x + 3}{6x - 4}\right)\)
Substitute the expression into \(f\):
\((fof)(x) = \frac{4\left(\frac{4x + 3}{6x - 4}\right) + 3}{6\left(\frac{4x + 3}{6x - 4}\right) - 4}\)
Multiply numerator and denominator by \((6x - 4)\) to simplify:
\((fof)(x) = \frac{4(4x + 3) + 3(6x - 4)}{6(4x + 3) - 4(6x - 4)}\)
Expand both numerator and denominator:
Numerator \(= 16x + 12 + 18x - 12 = 34x\)
Denominator \(= 24x + 18 - 24x + 16 = 34\)
\((fof)(x) = \frac{34x}{34} = x\)
Step 2: Find the inverse of \(f\).
Since \((fof)(x) = x\), it implies that \(f\left(f(x)\right) = x\), which means \(f\) is its own inverse function.
Therefore, \(f^{-1}(x) = f(x) = \frac{4x + 3}{6x - 4}\).
In simple words: When a function composed with itself gives back \(x\) (\((fof)(x) = x\)), the function is its own inverse.

Exam Tip: If \((fof)(x) = x\) holds true for all \(x\) in the domain, you can immediately conclude that \(f^{-1}(x) = f(x)\) without doing extra algebraic derivations.